Hints/Solutions for Homework 3

Transcription

1 Hints/Solutions for Homework 3 MATH 865 Fall 25 Q Let g : and h : be bounded and non-decreasing functions Prove that, for any rv X, [Hint: consider an independent copy Y of X] ov(g(x), h(x)) Solution: Since g, h are non-decreasing functions, (g(x) g(y))(h(x) h(y)) for all x, y Let X, Y be iid random variables Then we have Now, E[(g(X) g(y ))(h(x) h(y ))] E[(g(X) g(y ))(h(x) h(y ))] E[g(X)h(X)] E[g(X)h(Y )] E[g(Y )h(x)] Eg(X)h(Y ) + E[g(Y )h(y )] 2 ( E[g(X)h(X)] E[g(X)]E[h(X)] ) 2ov(g(X), h(x)) Q2 Prove that the following are equivalent (a) X n X in probability (b) There exist ɛ n such that P( X n X > ɛ n ) ɛ n (c) E min( X n X, ) Solution: (a) (b) For each k N, find N k > N k such that P( X n X > k ) k for all n N k Actually, we can take N Now, set ɛ n k if N k n < N k learly, ɛ n and P( X n X > ɛ n ) ɛ n (b) (c) Let ɛ n be as in (b) min( X n X, ) min( X n X, ) Xn X ɛ n + min( X n X, ) Xn X >ɛ n min(ɛ n, ) + Xn X >ɛ n Taking expectation, we get the inequality (c) (a) For ɛ (, ), E min( X n X, ) min(ɛ n, ) + P( X n X > ɛ n ) min(ɛ n, ) + ɛ n P( X n X > ɛ) P(min( X n X, ) > ɛ) E min( X n X, ) ɛ where the last inequality follows from the Markov s inequality Hence, X n p X, Q3 For s >, define the zeta function by n s n

2 Fix s >, and let X be a random variable with P(X n) n s / Show that the events are independent and deduce Euler s formula Prove also that {p divides X} p is prime p: ( p s ) P(X is square-free) ζ(2s) (Bonus) Let Y be an independent copy of X and let G be the greatest common divisor of X and Y Show that P(G n) n 2s ζ(2s) [ X is square-free No perfect square other than divides X] Solution: Let p, p 2,, p k be distinct primes P(p i divides X, i, 2,, k) P(p p 2 p k divides X) P(X mp p 2 p k ) m (mp p 2 p k ) s / p s p s m Letting k in above, we get P(p i divides X) p s i Therefore, the events {p divides X} p is prime are independent k P(X ) P(X is not divisible by any prime ) lim P(X is not divisible by some prime p, p n) n lim P(X is not divisible by p) n :p n P(X is not divisible by p) ( p s ) Easy to check, by the same argument, that the events {p 2 divides X} p is prime are independent Hence, P(X is square-free) P(X is not divisible by p 2 for any prime p) P(X is not divisible by p 2 ) ( ) p 2s ζ(2s), where the last equality follows from Euler s formula 2

3 P(G n) a,b:gcd(a,b)n c,d:gcd(c,d) c,d:gcd(c,d) n 2s P(X a)p(y b) P(X cn)p(y dn) (cn) s c,d:gcd(c,d) (dn) s (c) s Summing over n N, we have A n n 2s which yields A Q4 onsider the probability space (d) s An 2s (say) ζ(2s) ( ) (, ], B (,], λ where λ is the uniform measure on (, ] For any ω (, ], consider its unique binary representation ω ω ω 2 ω 3, where we forbid infinite sequence of s Define random variables X, X 2,, on (, ] by setting X k (ω) ω k Show that X, X 2, are iid Ber( 2 ) [You can use Durrett exercise 25] Solution: We need to prove that X, X 2,, X n are independent for each n N Fix ɛ, ɛ 2,, ɛ n {, } Let γ n i ɛ i2 i P(X ɛ,, X n ɛ n ) λ({ω (, ] : ω ɛ,, ω n ɛ n }) λ(γ, γ + 2 n ] 2 n From this joint probabilities, we can also compute marginals, P(X i ɛ i ) P(X ɛ,, X i ɛ i ) Therefore, ɛ j {,}, j<i ɛ j {,}, j<i P(X ɛ,, X n ɛ n ) 2 i 2 i 2 i 2 n i By Durrett exercise 25, X, X 2,, X n are independent Q5 Show that for any p > and for any random variable X, E[X p ] pt p P(X > t)dt P(X i ɛ i ) 2 n [Hint: Use Fubini We have proved Fubini s theorem for the product probability measures µ ν The similar proof also shows that the Fubini s theorem holds for the product measures µ ν when µ, ν are σ-finite measures (see Durrett (page 4) for the definition) The Lebesgue measure on is a natural example of σ-finite measure] [omment: For p, it gives a nice formula for the expectation of nonnegative random variable X as E[X] P(X > t)dt Durrett exercise 226(i) also gives a similar useful formula for E[X] when X is integer-valued] Solution: It s exactly Lemma 228 of Durrett s book (if you haven t figured that out yet, you should go over the textbook more carefully!) 3

4 Q6 Durrett 74 Solution: By Fubini s theorem for µ Leb, (F (x + c) F (x))dx (x,x+c] (y)µ(dy)dx cµ(dy) cµ() Q7 Durrett 25 [y c,y) (x)dxµ(dy) Let A i be the collection of the sets of the form {X i x i }, where x i S i In addition, let A i contains and Ω learly, A i is a π-system and σ(x i ) σ(a i ) To check X,, X n are independent, we need to verify that for each A i A i, i, 2,, n, If none of A i s are or Ω, then () follows from the hypothesis P(A A n ) P(A ) P(A n ) () P(X x,, X n x n ) P(X x ) P(X n x n ) x i S i (2) If one of A i is, then () holds trivially (both sides are equal to zero) If a few of the A i s are Ω, then we can still verify () by summing (2) over all x i S i for those coordinates i such that A i Ω Q8 Durrett 28 Solution: (i) P(X + Y ) x+y (x, y)d(µ ν) 2 µ({ y})ν(dy) x y (x)µ(dx)ν(dy) Since S : {y : µ({ y}) > } is countably many y, we can write the function y µ({ y}) as the infinite sum of indicator functions s S ysµ({ s}) Hence, by MT, µ({ y})ν(dy) ys µ({ s})ν(dy) µ({ s})ν({s}) s S s S (ii) Since X has continuous distribution, µ({x}) for all x Hence, P(X + Y ) µ({ s})ν({s}) by part (i) s S Q9 Durrett 223 Solution: Note that f(u ), f(u 2 ), are iid random variables with E f(u ) f(x) dx < Hence, by Theorem 229 of Durrett (or SLLN), I n : n n i Note that E[I n ] I By hebyshev s inequality, f(u i ) p Ef(U ) f(x)dx : I P( I n I > an /2 ) Var(I n) na 2 nvar(f(u )) na 2 a 2 E[f(U ) 2 ] a 2 f(x) 2 dx Q Durrett 224 (i) E X i k k 2 log k k log k + 4

5 (ii) We apply Durrett Theorem 227 First we check that mp ( X m) m km k 2 log k m m log m as m Therefore, n p S n µ n, where µn E[X X n] We claim that µ n µ from which the solution follows Note that n n µ n ( ) k k ( ) k k 2 log k k log k Let ϕ(x) x log x Then ϕ (x) x 2 log x x 2 (log x) 2 Thus, by mean-value theorem, we can estimate The claim follows ϕ(k) ϕ(k + ) sup ϕ (t) O(k 2 ) t [k,k+] 5